Question

A 200 gallon tank initially contains 100 gallons of water with 20 pounds of salt. A salt solution
with 1/4 pound of salt per gallon is added to the tank at 4 gal/min, and the resulting mixture is
drained out at 2 gal/min. Find the quantity of salt in the tank as it’s about to overflow.

Answer 1

The amount of salt \(A\) in the tank at time \(t\) is described by the differential equation,
\[A'(t)=\left(\frac{1}{4}\frac{\text{lb}}{\text{gal}}\right)\left(4\frac{\text{gal}}{\text{min}}\right)-\left(\frac{A(t)}{100+(4-2)t}\frac{\text{lb}}{\text{gal}}\right)\left(\frac{2\text{gal}}{\text{min}}\right)\]
which is simplified to the (linear) equation,
\[A'(t)+\frac{A(t)}{50+t}=1\]
Integrating factor:
\[\begin{align*}\log\mu(t)&=\int\frac{dt}{50+t}\\\\
&=\log|50+t|\\\\
\mu(t)&=50+t
\end{align*}\]
Distribute the IF:
\[\begin{align*}(50+t)A'(t)+A(t)&=50+t\\\\
\frac{d}{dt}[(50+t)A(t)]&=50+t\\\\
(50+t)A(t)&=\int(50+t)~dt\\\\
A(t)&=\frac{50t+\frac{1}{2}t^2+C}{50+t}\end{align*}\]
You're given the initial condition that \(A(0)=20\). Use this to solve for \(C\).

Next you'll want to find the time it takes for the tank to overflow. At any time \(t\), you have that the volume of water in the tank is
\[V(t)=100+(4-2)t\]
and so the tank will reach capacity for \(t\) that gives \(V(t)=200\). Find this \(t\).

Finally, plug in the overflow time and solve for \(A\).